Various Proofs for Irrationality of sqrt2

We all know the standard proof that the square root of two is irrational, and it's easily extended to all integers that are not perfect squares, but It just striked me yesterday that I have only seen one proof (which really is enough, but still =]).

One of the lecturers at the University of Sydney (I went there for work experience) showed me a new proof, which is from the perspective of set theory (it uses the fact a subset of the positive integers must have a smallest element), rather than the standard number theory results of divisibility. I thought some of you would find it interesting, I certainly did =]
--------------------------
Proof by Contradiction:

The statement that the square root of 2 is irrational is equivalent to the statement that a subset of the natural numbers [tex]S= \{ n \in \mathbb{Z}^{+} | \sqrt{2}n \in \mathbb{Z}^{+} \}[/tex] is not empty.

This set must have a least member, let it be u.
Let [itex]w=(\sqrt{2}-1)u[/itex]
Then [itex]\sqrt{2}w= 2u - \sqrt{2} u[/itex].

Since u is a member of S, sqrt2*u is a positive integer, so sqrt2*w is also a positive integer. Therefore w is also a member of S.

However, sqrt2 minus 1 is between 0 and 1, so w < u. Yet u is the least member of S.

We all know the standard proof that the square root of two is irrational, and it's easily extended to all integers that are not perfect squares, but It just striked me yesterday that I have only seen one proof (which really is enough, but still =]).

One of the lecturers at the University of Sydney (I went there for work experience) showed me a new proof, which is from the perspective of set theory (it uses the fact a subset of the positive integers must have a smallest element), rather than the standard number theory results of divisibility. I thought some of you would find it interesting, I certainly did =]
--------------------------
Proof by Contradiction:

The statement that the square root of 2 is irrational is equivalent to the statement that a subset of the natural numbers [tex]S= \left{ n \in \mathbb{Z}^{+} | \sqrt{2}n \in \mathbb{Z}^{+} \right}[/tex] is not empty. (For some reason the braces won't appear, damn).

Are you quite sure you don't mean "The statement that the square root of 2 is rational??
The braces don't appear because braces are part of the "control" symbols for LaTex. Use "\{" and "\}" instead.

This set must have a least member, let it be u.
Let [itex]w=(\sqrt{2}-1)u[/itex]
Then [itex]\sqrt{2}w= 2u - \sqrt{2} u[/itex].

Since u is a member of S, sqrt2*u is a positive integer, so sqrt2*w is also a positive integer. Therefore w is also a member of S.

However, sqrt2 minus 1 is between 0 and 1, so w < u. Yet u is the least member of S.

I'm no great hand at number theory but I remember having read the following more algebraically minded proof;
Let p/q be a positive fraction in its lowest terms such that (p/q)^2 = 2

then p^2 = 2q^2
From this,
(2q - p)^2 = 2(p-q)^2

again 2 = (2q - p)^2 / (p - q)^2
the same property of the fraction p/q
since q<p<2q gives p-q<q
therefore we have another fraction with a smaller denominator and this contradicts our assumption, neatly establishing the irrationality of the positive solution of the equation
x^2 - 2 = 0